Thanks to Randy of our CCNA program for this suggestion. Randy wanted some guidance on how to solve the subnetting questions in ICND1 and ICND2 very quickly. The ability to do this is often the difference between a passing score and a failed attempt.

WARNING: You must master subnetting using our course or some other trusted materials before you start using these shortcut approaches. It is a common issue for Cisco candidates to move directly to subnetting shortcuts for the exams without fully understanding exactly how subnetting functions.

For this series of posts, we will use simulated exam questions from ICND1 and ICND2. Well, with all that out of the way – let’s have some fun. You will find that once you “turn the corner” on subnetting, you will pray for many of these questions in the exam. It is an opportunity to solve questions quickly and be 100% convinced that your response is “spot on”.

ICND1 (CCENT)

Question 1: What is the last usable address in the subnet of a host with the address 192.168.1.134 and the subnet mask of 255.255.255.240?

Step 1: Since the is the first subnetting question I have encountered in my exam, I am going to use this as my opportunity to build my Powers of Two reference chart on my scratch paper.

2^7=128 | 2^6=64 | 2^5=32 | 2^4=16 | 2^3=8 | 2^2-=4 | 2 ^1=2 | 2^0=1

Step 2: 192 in the first octet tells me I have a Class C address. I memorized these number ranges; some students like to list those on the scratch paper as well. The default subnet mask for the Class C space is 255.255.255.0. You also need to memorize these, and again, many students like to list these on scratch paper as well.

Step 4: I go four bits deep (from left to right) in the Power of Two chart. This tells me that the subnets increment on 16:

192.168.1.16…32…48…64…80…96…112…128…144

Step 5: We can see that this host lives on the 192.168.1.128 subnet. The broadcast address for this subnet is one less than the next subnet of 144, so that is 143. The last usable is 142. Our answer – 192.168.1.142. Wooohooo! Bring on more of these!

Did you have an even easier way to arrive at the answer? Let us know in the comments.

When I analyze 254 as the “interesting” octet, several things come to mind.

1. 256 – 254 = 2 = calculated max hosts = power of two of the the single host bit
2. Network bits in this octet (8 – the single bit) = seven
3. Often found as the third octet of a Class B network that requires address-space for more than 256 hosts.

When I analyze 252 as the “interesting” octet, several things come to mind.

4. 256 – 252 = 4 = calculated max hosts = power of two of the the 2 host bits
5. Network bits in this octet (8 – the two host bits) = six
6. Efficient for point to point connections, no wasted addresses

When I analyze 248 as the “interesting” octet, several things come to mind.

7. 256 – 248 = 8 = calculated max hosts = power of two of the the 3 host bits
8. Network bits in this octet (8 – the three host bits) = five

When I analyze 240 as the “interesting” octet, several things come to mind.

9. 256 – 240 = 16 = calculated max hosts = power of two of the the 4 host bits
10. Network bits in this octet (8 – the four host bits) = four

When I analyze 224 as the “interesting” octet, several things come to mind.

11. 256 – 224 = 32 = calculated max hosts = power of two of the the 5 host bits
12. Network bits in this octet (8 – the five host bits) = three

When I analyze 192 as the “interesting” octet, several things come to mind.

13. 256 – 192 = 64 = calculated max hosts = power of two of the the 6 host bits
14. Network bits in this octet (8 – the six host bits) = two

When I analyze 128 as the “interesting” octet, several things come to mind.

1. Octets like 252 and 254 can cause confusion when they are the third octet.
2. FLASHCARD the fact that networks start on even, yes ZERO is considered EVEN, boundaries.
3. The last usable address for either 172.16.0.1/23 or 172.16.1.2/23 is 172.16.1.254
4. The last usable address for either 172.16.2.3/23 or 172.16.3.4/23 is 172.16.3.254

When the exam gives us a 255.255.255.240 mask I usually subtract the interesting octet value from 256. So 256 – 240 gives us the network multiple of 16. I like the method above when the exam gives us the prefix length. 192.168.1.134 / 28, means they used 4 bits for subnetting, 28 – 24 =4.
So, 4 deep into the values is 16. These are the only 2 shortcuts that I learned (there are hundreds, I’m sure) and they work quickly.

My way, well read from ccna book when I studied few years ago, is to get the block (256-240) which is 16. Then I double them, 32,64,128 until getting closer to 134. If I add another 16 on to 128, it’s going to be 144, the next subnet. I then know it range is between 128 and 143. The usable ip is in between.

(hopefully that’s readable)
The first row are your powers of two. Subtract one to get a Wildcard, Subtract two for the number of usable hosts.

The second row is your SN Mask and is formed by adding the previous to the row above.

The last row is the number of bits (e.g. /25 through /32). If you’re working in the 2nd or 3rd octect, it’s fairly easy to shift left.

To solve the problem, the mask of …240 corresponds to 16 addresses. 16 goes into 134 eight times… 16 * 8 = 128, so the *first* usable address is 129 (128 being the network). 16*9 = 144, so the *last* usable address is 142 (143 being the broadcast).

This is essentially the same process, but I thought I’d share my chart — it helps to have it visible during testing when your internal calculator can screw you up… eventually you don’t really need the chart in practice.

192.168.1.134 255.255.255.240
here we have subnet=256-240=16
hence we have the available subnet as 0,16,32,……,128,144,…,224,240
and we have the magic number in the fourth octect and that is between
128 and 144 and address range between the 128 and 144 is 129-142
as we can’t assign the first and the last ip address to a device hence the last usable address is 192.168.1.142

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